Chapter 7 - Section 7.3 - Double-Angle, Half-Angle, and Product-Sum Formulas - 7.3 Exercises - Page 561: 11

$\frac{3}{8}-\frac{1}{2}cos2x+\frac{1}{8}cos4x$

Use $sin^2x=\frac{1-cos2x}{2}, cos^2x=\frac{1+cos2x}{2},$, we have $sin^4x=(\frac{1-cos2x}{2})^2=\frac{1}{4}(1-2cos2x+cos^22x) =\frac{1}{4}(1-2cos2x+\frac{1+cos4x}{2})=\frac{3-4cos2x+cos4x}{8}=\frac{3}{8}-\frac{1}{2}cos2x+\frac{1}{8}cos4x$